Cho 40 gam hỗn hợp X gồm Fe3O4

B

- Hướng dẫn giải

\(\begin{gathered} + \,BTE:\,\,{n_{F{e_3}{O_4}}} = {n_{Cu\,\,pu}} = x \Rightarrow 232x + 64x = 40 - 16,32 \Rightarrow x = 0,08. \hfill \\ + \,\,\left\{ \begin{gathered} FeC{l_2}:\,\,0,24\,\,mol \hfill \\ CuC{l_2}:\,\,0,08\,\,mol \hfill \\ \end{gathered} \right\}\xrightarrow{{AgN{O_3}\,\,du}}\left\{ \begin{gathered} Ag \downarrow :\,\,0,24\,\,mol \hfill \\ AgCl \downarrow :\,\,0,64\,\,mol \hfill \\ \end{gathered} \right\} \Rightarrow {m_{ket\,tua}} = 117,76 \approx \,\boxed{118} \hfill \\ \end{gathered} \)